Blue Cab operates 15% of the taxis in a certain city, and Green Cab operates the other 85%. After a nighttime hit-and-run accident involving a taxi, an eyewitness said the vehicle was blue. Suppose, though, that under night vision conditions, only 80% of individuals can correctly distinguish between a blue and a green vehicle. What is the (posterior) probability that the taxi at fault was blue? In answering, be sure to indicate which probability rules you are using. [Hint: A tree diagram might help. Note: This is based on an actual incident.]

0.4138 or 41.38%

Step-by-step explanation:

Prob(a) = 0.15

Prob(b|a) = 0.80

A = event that taxi is blue

B = witness says it is blue

Prob(taxi is not blue) = 1 - prob(a)

= 1- 0.15

= 0.85

Prob(b|a') = 1 - prob(b|a)

= 1 - 0.8

= 0.20

Using addition rule

P(b) = p(a)p(b|a) + p(a')p(b|a')

= 0.15 * 0.8 + 0.85 * 0.20

= 0.29

Now we are to find probability that the taxi that is at fault is blue

According to bayes theorem

P(a|b) = P(a)*p(b|a) / p(b)

= 0.15 x 0.8 / 0.29

= 0.4138

constant of proportionality = 3 and the equation is c = 3p.

step-by-step explanation:

c = kp   where c = amount of caffeine in mg., p = oz of pepsi and k is the constant of proportionality.

substituting the values given:

24 = 8k

k = 3.

the same value of k will be obtained if we substitute the other given values.